Question
THE 14TH TERM OF AP IS 96 WHILE THE 25TH TERM IS 173 FIND THE
A.19TH TERM
B.SUM OF 13TH AND 56TH TERM
A.19TH TERM
B.SUM OF 13TH AND 56TH TERM
Answers
TN=A+(N-1)D
T19=5+(19-1)7=5+18(7)
=15+126=T19=131
T19=5+(19-1)7=5+18(7)
=15+126=T19=131
http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
Xn = a +(n-1)d
96 = a + 13 d
173 = a + 24 d
---------------
-77 = -11 d
d = 7
then a = 96 - 13*7 = 5
so
X19 = 5 + 18*7 = 131 etc
Xn = a +(n-1)d
96 = a + 13 d
173 = a + 24 d
---------------
-77 = -11 d
d = 7
then a = 96 - 13*7 = 5
so
X19 = 5 + 18*7 = 131 etc
LOL, yes :) good
173=a+24d
96=a+13d
77=11d
d=7
96=a+13×7
96=a+91
a=5
19th term = a+18d
5+18×7
=131
96=a+13d
77=11d
d=7
96=a+13×7
96=a+91
a=5
19th term = a+18d
5+18×7
=131
The14th term of an ap is 96 while the 25th term is 173 find the (a)the19th term (b)sum of 13th and 5th term (c)product of
Maths
Nothing
The forth term
Tn=a+(n-1)d
93=a+(14-1)d
96=a+13d---eqn(i)
173=a+(25-1)d
173=a+24d---eqn(ii)
a+13d=96
a+24d=173
11d=77
d=7 a+13(7)=96
a+91=96
a=96-1
a=5
(a) tn=5+(19-1)7
T19=5+(18)7
T19=5+126
T19=131
93=a+(14-1)d
96=a+13d---eqn(i)
173=a+(25-1)d
173=a+24d---eqn(ii)
a+13d=96
a+24d=173
11d=77
d=7 a+13(7)=96
a+91=96
a=96-1
a=5
(a) tn=5+(19-1)7
T19=5+(18)7
T19=5+126
T19=131
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What about b the sum of 13th and 56th term
I want to learn maths
Pls can i learn maths
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I need answers
What about b the sum of 13th and 56th term
Am looking for the sum of 13th and 56th term
The product of 6th and 13th term
The product of 6th and 13th term
Good job
(C). Tn=t1+(n-1)d
T6=5+(6-1)7
T6=5+5×7
T6=5+35
T6=40
T6=5+(6-1)7
T6=5+5×7
T6=5+35
T6=40